71 I
ANALYTICAL DYNAMICS
BEING A
SYNOPSIS OF LEADING TOPICS
-— IN THE ---
Analytical Theory of Dynamics with
Numerous Examples and Selections
from  Newton's Principia
and Other Sources.
-BY ---
ARTHUR S. HATHAWAY,
Professor of Mathematics, Rose Polytechnic Institute,
TERRE HAUTE, INDIANA.
1906 


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ANALYTICAL DYNAMICS
Being a Synopsis of Leading Topics in the Analytical Theory
of Dynamics with numerous Examples and Selections
from Newton's Principia and other Sources.
BY
ARTHUR S. HATHAWAY,


PROFESSOR OF MATHEMATICS, ROSE POLYTECHNIC INSTITUTE,
TERRE HAUTE, INDIANA.


1906.








Notes and Examples on Dynamics
KINEMATICS
We consider a point, moving on a given curve called its path.
At time t (fig. 1), its arc distance from a fixed point A will be arc
AP- s, at time t' it will be arc AP'= s', and so on, changing continuously as the time changes. In other words, s is a continuous
function of t, and s' is the same function of t'.
SPEED   AND VELOCITY
The average speed from  time t to time t', is the average
distance passed over per unit time, or the fraction, (s'-s):(t' —t)
The speed at time t is the limit of the above average speed.
(Understanding that t' approaches t as a limit). It is denoted by
v, and by definition,
(1)  v=ds/dt=Ds (D stands for d/dt, here and hereafter).
The speed is positive when the arc distance is increasing, and
negative when the arc distance is decreasing.
The average velocity from time t to time t' is the average displacement in space per unit time, in direction and magnitude. If
the point be at P (fig. 1) at time t, and at P' at time t', then the
average velocity is represented by the line PS' which is the chord
PP' extended in the ratio 1: (t'-t). On the contrary, the average
speed is represented by the arc PS" which is the arc PP' extended
on the path in the ratio 1: (t'-t).
The velocity at time t is the limit of the above average velocity.
Since the limit of a vanishing chord which is kept indefinitely produced, is the tangent line, the velocity at P is represented by a
line PS, on the tangent to the path at P.
The velocity PS determines the direction of motion at P, i. e.
the ultimate direction from P to a succeeding position P'. Also,
its numerical measure is the speed v. For the vanishing arc and
chord PP' have a limiting ratio of unity, so that the speed and
velocity, which are limits of proportionals to them, will be in that
ratio, as to length, or be equal lengths (the former measured properly on the path as a positive or negative distance, the latter on the
tangent, as a displacement in space, but both in the same direction




from P). Remembering that the velocity is tangential, it is therefore completely determined by the speed v (the path and position
P being known).
The term velocity as commonly used means either speed or
velocity, according to the context. For motion along a given path
the "compound" of two speeds is their ordinary sum, but velocities
derived from displacements combine by the parallelogram law of
displacements, and not by numerical addition, since they are displacements per unit time.
ACCELERATION-THE HODOGRAPH
The average acceleration from time t to time t' is the average
increase of velocity per unit time.
The acceleration at time t is the limit of the above average
acceleration.
From a fixed point 0 (fig. 1) draw lines OB, OH, OH', etc.,
representing, in direction and magnitude, the velocities of the
point at A, P, P', etc., (and so parallel to the tangents to the path
at those points, and of lengths representing the speeds at those
points, V, v, v', etc.) The curve B..H.. H'..., is called the
hodograph of the given motion.
Now HH' is the resultant of the velocities OH' and -OH i. e.,
the increase of velocity from time t to time t'; and if it be produced
to T' in the ratio 1: (t'-t), HT' will represent the above average
acceleration; and it will also represent the average velocity of the
corresponding point in the hodograph, between the same times.
Thus its limit HT, on the tangent to the hodograph at H, represents simultaneously the acceleration at P, and the velocity at H.
In words,
The acceleration of a moving point is the velocity of the corresponding point in its hodograph.
TANGENTIAL AND NORMAL ACCELERATIONS
In fig. 1, draw the perpendicular, TK, to OH produced. Then
HK and KT are called the tangential and normal accelerations, at
P, because they are the components of the whole acceleration HT,
parallel and perpendicular to the tangent at P. The normal ac'celeration KT, drawn from P, must lie towards the center of curvature, C, since it is in the direction in which the velocity PS begins
to turn as the moving point leaves P, or towards the concave side,of the path.
We proceed to find the measures of the tangential and normal
accelerations.
Lay off on the velocities OB,...OH,...OH'..., their corre

4




sponding positive units OL,.. OM,....O'..., i. e. such that OB=
V.OL, OH=v.OM, OH'-=v'.OM'.       This gives a curve LMM'
of unit radius, whose arc measures from L as origin are, arc
LM = —, arc LM'=-'.
These are the radian measures of the angles through which the
tangent to the path has turned, in the motion of its point of contact from A to P and to P'.
The curve LMM' (on a sphere of. unit radius and center 0),
may be called the angular directrix of the path, and ~, the angular
direction of the path at P.
The angular velocity of the direction of motion PS, is the
velocity of the corresponding point M on its angular directrix; and
is represented at time t by a tangent MN at M, whose numerical
measure is dq/dt, the angular speed.  (Remember that the treatment of velocity and speed on this curve is the same as on any
other, p being the arc distance of the moving point at time t.)
Draw on fig. 1, HH" parallel to MM' to intersect OH' in H".
Then,HHT' is the resultant of HH" and H"H', i.e. of v.MM' and
(v'-r) OM'. Hence, dividing by t'-t, to obtain HT' and its components, and thence, their limits,HT, and vMN, Dv.OM, the latter
must be KT, HK, respectively, or,
HK=-Dv.OMl, KT=vMN.
Remembering that the measure of OM is unity, and of MN,
d4/dt, we have:
(2) Tangential acceleration=f:=Dv-dv/dt.
(3) Normal acceleration =vdc/dt =v2do/ds
-v2/R, where R-ds/dc, the radius of curvature (PC) of the
path. (Calc. Art. 83.)
Note. The above is differentiation of the velocity OH, D being
a special differential symbol whose multiplying factor is n=1/(t' —t), so that Dtlim.n (t'-t)  1.
Thus let, OM —u, and denote resultant addition and subtraction by + and -.
Then,
D (vu) — [T=.n (v'u-' vu) =i —m.n (v'u' —vu' +vu' ---vu)
-lim. [n (v'-v).u'+nv (u'-u) ] — Dv.u+vDu.
and Du-=MN, the angular velocity.
MOTIONS IRRESPIECTIVE OF PATH
The character of a motion, irrespective of path, is determined
by the function which s is of t. This function may be given explicitly, or implicitly by sufficient equations between s and t and
other variables for the explicit solution of s in terms of t. From


5




such value of s we find at once by differentation, v==Ds, and f=
Dv==D2s. These may be written
(1) ds=vdt, (2) dv=fdt, from which, (4) vdv=fds.
Conversely, (1) and (2) constitute two independent equations
between the four variables t, s, v, f. Since one of these variables
is independent, there must be three independent equations between
them.  The third equation is sought in the form  s=-Ft.   (F"function of"); but it may be given by the problem as an equation
between all four variables or less. In such a case, integrations are
to be performed, which introduce arbitrary constants. These constants are to be determined by some assumed starting conditions,
i. e., an assigned value of s, and v, at a given time t (usually time
zero.) We outline methods in certain cases.
(a) When f is given in terms of t.
Find v in terms of t by integrating dv-fdt; thence find s
in terms of t by integrating ds-vdt.
(b) When f is given in terms of v.
Find t and s, each in terms of v, by integrating dt-dv/f,
ds-vdv/f.
(c) When f is given in terms of s.
Find v2, and thence v, in terms of s, by integrating
2vdv-2fds; thence find t in terms of s by integrating dt ds/v.
Since v  Ds, f-D2s, the most general analytical equation between the four variables is a differential equation of second order,
for s in terms of t. This is called the differential equation of the
motion. The equation for s in terms of t is the integral equation
of motion. We consider some types of motion, with their integral
and differential equations, leaving the verification as exercises.
Notice that the integral equation always contains two constants
not in the differential equation.  These are consequences of two
integrations.
TYPES OF MOTION
(a)  Unifornm Motion.   Equal distances are passed over in
equal times. sat-+b. D2s=0.
(b)  Uniformly Increasing Motion. Equal speeds are gained
in equal times. s   2at2+-bt+-c, D2s  a.
(c) Geometric Motion.   From   a given origin, called the
center, arc distances at successive equally different times are in
geometrical progression. s=a.exp(nt)+b, D2s —nDs.
Note exp (x)-that power of the naperian base e-2.71828 —
whose exponent is x. dexp(x)    exp(x).dx, and the same for D.
(d) Harmonic MVotion. The motion of a point on a diameter
of a' circle which is the orthogonal projection of a point moving
uniformly on the circumference.
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s-c=-acos(nt+b), or asin(nt+b'), or Acos(nt)+Bsin(nt)
[equivalent forms.]
D2s+n2 (s-c) O.
The center is s=c. The radius or amplitude is a==/ ( A2+B2)
The angular speed is n, the period, 27r/n. The phase angle is
b=b' —7r=tan- (B/A).
(e)  Hyperbolic Motion. The "harmonic" functions above
replaced by "hyperbolic" functions. This includes geometric motion, and is in general the compound (sum) of two geometric motions of reciprocal ratios, as in the second form.
s —c=Acosh(nt) +Bsinh(nt), or aexp(nt)+bexp(-nt) &c.
D2S=n2 (s-C).
(f) Harmonic Motion of Geometrically Decreasing Amplitude.
s-c-a exp (1 —  t).cos(nt+b).
D2s+kDs+ (n2+1/4k2) (s-c)=O.
Note.  To prove that the integral equation includes every
solution  of the differential equation, let s=x.exp( —lkt) +c,
whence x is any solution of D2x-+n2x-O.
FIELDS AND     MEDIUMS
In nature, a moving point is associated with matter, and because of the inertia property of matter, cannot be conceived as in
itself changing its state of rest or of uniform motion in a straight
line. In this view of the motion of a point, it will be called a
particle. The motion of a particle is simply the geometrical motion of a point, and its velocity and acceleration are geometrical
quantities only, but we regard the acceleration, or change of
velocity per-unit time, as being produced by some cause outside the
particle.
Thus, a particle near the surface of the earth is given an
acceleration g, downward, wherever it is placed, whose effect is only
anulled by some other cause which gives the particle the opposite
acceleration, so that the resultant acceleration of the particle is
zero.
A space in which a particle is given an acceleration which
varies only as its position varies, will be called a field. In a field
the acceleration, in direction and magnitude, is the same at the
same point at all times and becomes the acceleration of the particle
when it arrives at that point.
In a constant field, the acceleration is the same in magnitude
and direction at all points of the field. It will be suggestive to call
its direction downward, and to illustrate the magnitude, g, by the
numerical value 32, or 981, in the foot second, or centimeter
second system.


7




In a central field, the acceleration is at all points either towards or from a fixed point called the center of the field. Practically the earth's field is such in which the acceleration is towards
the center of the earth and inversely as the square of the distance
from the center. When the acceleration depends upon the distance
from the center only, the absolute acceleration is the acceleration
at units distance from the center.
We may consider a uniform medium, which retards speed, but
produces no change in the direction of motion. Its acceleration is
therefore wholly tangential, and of opposite sign to the speed; it
should be in magnitude, some function of the speed alone, which
vanishes with the speed, and increases as the magnitude of the
speed increases. The absolute retardation of a medium is its retardation for unit speed, and is considered a function of the density
of the medium which increases with increasing density. It is often
assumed proportional to the density of the medium.
CONSTRAINED PATHS. FRICTION.
A particle in any field may describe any path, regarded as a
tube in which the particle is constrained to move. The tube serves,
in effect, as a producer of acceleration in addition to the field, since
its confining walls change the direction of the velocity, and that is
acceleration when measured'per unit time. We generally consider
that this acceleration is wholly normal, i.e., not affecting the speed.
Also, the side of the tube which produces the acceleration by
"pressing" upon the particle is that through which the particle
would pass, if the field alone acted. All other sides but the one
exerting pressure may be removed without affecting the motion on
the path.
The normal acceleration produced by the pressure of a tube
combines with the normal acceleration of the field to produce the
normal acceleration of the particle (v2/R); i. e., it is the resultant
of the normal acceleration v2/R and the reversed normal acceleration of the field.
The tangential acceleration which we conceive the friction of a
rough path to produce, is always opposed to the motion. It is
treated as constant, or in a constant ratio to the normal acceleration of the path, called the coefficient of friction. The whole retardation of friction is called into play only on a particle in motion,
or on the point of moving.
EXAMPLES I.
1. If the tangential acceleration is always zero, the motion
is uniform; and conversely. If the normal acceleration is always


8




zero, the path is a straight line; and conversely. If the whole
acceleration is always zero, the point moves uniformly in a straight
line; and conversely.
2. If a particle move freely in a field of constant acceleration,
show that the hodograph motion is uniform motion in a straight
line in the direction of the field.
3. A shot is fired at a velocity of 320 feet per second at an
inclination whose slope is 4/3. Draw the hodograph of the subsequent motion of the shot to a scale in which an inch represents 64
units of speed, and determine by it the velocity at the end of successive seconds, for the first sixteen seconds. Show that the greatest height is reached at the end of eight seconds.
4. Figure 1 is the elliptic path of a planet, P about the sun
at the focus F; and its hodograph is a circle. Sketch on the figure
the velocity of the planet at various points, including the ends of
the axes and focal ordinates. Also sketch the direction of the
acceleration of the planet at each point.
5. Show that in the speed-curve of any motion the space described between the times of two abscissas is represented by the
area of the curve between the corresponding ordinates. Also, that
the tangential acceleration at the time of any abscissa is represented by the slope of the curve at the point corresponding to that
abscissa.
Note. The speed-curve of a motion is the locus of a point in
the coordinate plane, whose abscissa represents the time, and ordinate, the speed. It furnishes a representation of all four
variables, t the abscissa, v the ordinate, s the ordinate area, f the
slope.
6. Show that the speed-curve of constant tangential acceleration is a straight line whose slope is the acceleration, and
y-intercept, the speed at time zero.
7. Draw    the  speed-curve  of  the  semi-cubical motion
s-_8t /2.
8. A point begins with speed V, and moves with constant
tangential acceleration g. If s be the distance passed over, and v,
the speed acquired, at the end of t units of time, show by the calculus, and by the speed-curve, that
v-gt+V, s ---gt2+Vt, v2-V2-      2gs.
9. Show that with constant tangential acceleration, the average speed is the arithmetical mean of the initial and final speeds.
10. Find the motion whose average speed is the geometric
mean of the initial and final speeds.
Note. If a be the initial speed, s- =(av) t. Squaring and
putting ds/dt for v, this becomes ads/s'-dt /t2.
Ans. s=at/(1+ct)


9




11. Find the space passed over in any given time by a particle
moving with a speed which varies as the square of the time.
12. Find the space described from rest at any time by a point
whose acceleration varies as the n' th the power of the time.
13. Find the speed acquired from rest, when a particle is
accelerated towards a center proportional to the distance. Find
the time of describing a given space, and the time to reach the,
center.
14. If the tangential acceleration of a motion vary inversely
as the speed, the square of the speed varies uniformly with the
time, and the cube of the speed, uniformly with the arc distance.
15. Determine the motion when the space described in any
time, varies as the speed acquired.
16. Show that the compound of two harmonic motions
of the same period is a harmonic motion of that period.
17. Show the following construction for the distance of a.
point in harmonic motion from its center at any time.
Draw an initial line OX, an angle A OX, minus the phase
angle b, on which OA=a, the amplitude, and describe the circle on
OA as diameter. Then a line coinciding with OX at time zero and
revolving about 0, uniformly in the period of the motion, cuts this
circle at any time in a point P such that OP is the required distancefrom the center at that time. OA is called the vector of the harmonic motion.
18. OA and OB are the vectors of two harmonic motions of
the same period. Show that the vector of their compound is OC,.
where OACB is a parallelogram.
PROBLEMS OF A CONSTANT FIELD
19. If a particle is constrained to move on a line inclined at
an angle i to the horizontal, show that its tangential acceleration
is gsin i down the line, and that the constraint of the line must
produce a normal acceleration gcosi. The line must lie immediately beneath, not above the particle.
20. Show that the time of descent of a particle down any
chord of a vertical circle, beginning at the highest point, is constant, and the same as the time of descent down any chord terminating in the lowest point of the circle.
21. The right line of quickest descent in a vertical plane from
a given point to the circumference of a lower circle in the plane,
passes through the lowest point of the circle. The right line of
quickest descent from the circumference to a lower point in its
plane, passes through its highest point.
22. Find the right line of quickest descent between two,
circles in the same vertical plane.


10




23. If the projection of a right line on a vertical line be
constant, show that the time of descent on that line varies as the
length of the line, and that the speed acquired is constant.
24. A plane is inclined at an angle i to the horizon, and contains a line that makes an angle a with the intersection of the plane
with the horizon. Prove that the acceleration down this line is
gsini sina.
25. A particle moves in a circle in a vertical plane. Prove
that its tangential acceleration is g.sino, where 4 is the angle between the vertical and the radius to the point. Thence express f
in terms of s, measured from the lowest point of the circle, the
radius being 1. Ans. f —g.sin(s/l)
26. If in the preceding example, the particle start from rest
at s=a, find v in terms of s, and express t as a definite integral between the limits a and s.
27. The Simple Pendulum. If in the preceding example,
the arc described be small, show that the motion is practically the
harmonic motion s-a.cosnt, where n2 —g/l. The period is consequently 27rV/ (l/g.)
Note. For small arcs sin(s/l)=-s/l, practically.
28. Calculate the length of a second's pendulum  (period 2
seconds.)
29. A pendulum beating seconds is 39.14 inches in length.
The bob is screwed up one turn, the screw being 32 threads to the
inch. Show that it will gain 34.7 seconds in a day in consequence.
30. A particle moves on a given curve (freely or by constraint.) Its arc distance from a fixed point being s, and its height
above a given horizontal plane being y, at time t, show that the
tangential acceleration is f — gdy/ds. Hence from   (4), if its
height is 1t, and speed V, at the start, then at time t,v2 -V2+-2g(h-y). In particular if the particle fall vertically from
rest V2_2-g(h-y), where h-y is the distance fallen.
31. Show from the result of the preceding example, that a
particle in motion on any path has at any time the same speed that
it would acquire by falling vertically from a fixed horizontal plane
to its given position.
Note. Let k be the height of a fixed horizontal plane above
the starting position such that 2gkc=7V2. Then v2  2g(k-+h —y)
and k-+h-y is the height of that plane above the particle. That
plane is the level of zero speed for the motion, and the particle cannot rise above it, since v2 must be a positive quantity. The path
may not rise to that level, in which case the speed cannot become
zero.
32. Consider a cycloid generated by a point on a circle in a


11




vertical plane, which rolls on its higher horizontal tangent. With
horizontal and vertical axes of x and y, through the lowest point,
show that its equation is,
x= —a(2 —sin2,), y=a(1-cos2O)
where a is the radius of the generating circle, and 20 is the angle
which the diameter through the tracing point (x,y) has turned
from its initial position at the vertex.
Also, s being its arc from the vertex, show that ds-4acospd+,
s=4asino, or s'-8ay.
Also, dy/dx-tanm, whence < is the angle through which the
tangent to the cycloid has turned from the vertex, and ds/dq is
the radius of curvature at the point (x,y).
33. If a particle is constrained to move in the above cycloid,
show that its motion is harmonic, with the vertex as center, and
angular velocity, -X/ (g/a). (Use s2-8ay, and ex. 30). Deduce
thence, the result of ex. 18, by assuming the small circular arc
practically a small arc of a cycloid of the same curvature as the
circle at its lowest point.
34. Find the speed with which the above circle must roll on
its apper tangent in order that a point on its circumference may
execute the motion of a particle sliding on the cycloid which it
generates.
35. In the preceding motion of a particle, show that the constraint of the cycloidal path produces an acceleration of 2gcosq
towards the point of contact of the rolling circle.
36. The conical pendulum. A particle is on the inner surface of a smooth sphere, the horizontal small circle through its
position being I units below the center of the sphere, and having a
radius R. Show that if it is started in this small circle with speed
v-=R/ (g/l), and for that speed only, it will describe the small
circle as a path, and with uniform motion whose time of revolution is the period of a simple pendulum of length 1.
Note. Consider the small circle to be a smooth tube. A particle
started in it with any speed will move uniformly, since there is no
charg:e of level (ex 31).  The resultant constraint at any point
must produce, an acceleration v2/R towards the center of the circle,
and an upward acceleration balancing gravity. If this resultant is
towards the center of the sphere, the tube may be removed without
affecting the motion, because the sphere alone will produce the
needed restraint.
37. What is the least horizontal velocity with which the above
particle may be started at the lowest point of the sphere so that it
will describe a vertical great circle.
Ans. With speed due to the zero level 3/2 the radius above
the center.


12




38. If in the above example, the zero level of the starting
speed is.n times the radius above the center, where n is less than
3/2, show that the particle will leave the spherical surface at an
arc distance from the top whose radian measure is cos-1(2n/3).
39. A particle on the outside of a smooth sphere, at its highest point, is started horizontally with a speed whose zero level is n
times the radius above the center. Show that if n-3/2 or greater,
the particle leaves the sphere at once, and that if n is less than 3/2,
it leaves the sphere at a level which is 2n/3 above the center.
40. If a particle is placed at any point on the upper half of a
smooth sphere, through what arc will it slide before leaving the
surface?
41. A particle is constrained to lie on a smooth cycloid with
vertex upward, x=-a(2(-+sin2)), y=a(cos20-1). The level of
zero speed is na above the vertex. Find the direction and magnitude of the acceleration produced by the constraint of the path at
any point (x,y).
42. Two particles are projected from the same point, in the
same direction, and with the same speed, but at different instants,
in a smooth circular tube whose plane is vertical. Prove that the
line joining them always touches a circle whose radical axis with
the given circular path is the level of zero speed.
Note. Draw two circles 0. 0' whose radical axis is horizontal; draw the tangent to 0' at any point T, and let it intersect
0 in P, Q, and the radical axis in R. Draw the perpendiculars
PL, QMl, to the radical axis, and show that TP2:TQ —PL:QM.
Also, as T moves on 0', show by successive positions that the speeds
of PQ are as TP:TQ. Hence show that if P moves with the speed
due to the zero level iILR, Q does the same.
43. A particle hangs freely from a fixed point by an inextensible string 2 feet in length. It is projected in a horizontal
direction with a speed of 20 feet per second. Show that the accelerations produced by the -string at the highest and lowest points
are as 29 to 5.
44.  Given a smooth parabolic tube, convex side up and axis
vertical (x' —4ay). A particle is started from the vertex in this
tube with given speed. Show that if the level (h) of zero speed of
the particle is below the directrix, the pressure of the tube is always
outZvards, so that the outer half may be removed without affecting
the motion; that if this zero level is above the directrix the pressure is always inwards, so that the inner half may be removed
without affecting the motion; and that if the directrix is the zero
level, the pressure of the tube is always zero, so that the particle
will describe the path freely.


13




Note. The normal acceleration of gravity at (x,y) is gdx/ds
inwards, and tan  = —dy/dx.    Prove the tube exerts the inward
normal acceleration Vag (h-a) /V (a-y) 3, at the point (x,y),
etc.
45. A heavy particle hangs freely from   a fixed point by a
fine elastic wire. The wire produces an acceleration upward proportional to its extension of length and the extension for which the
particle hangs at rest is I units. If the particle be pulled down a
units, and released, show that it describes harmonic motion of
amplitude a, and that its period is the same as a simple pendulum
of length 1 units.
46. A horizontal platform vibrates up and down harmonically
in a period of one second. Find its greatest amplitude in centimeters such that particles may rest upon it undisturbed.
CENTRAL FIELDS
47. A    partical  starts  from  rest  at  a   given  point
in a central field whose acceleration towards    the  center  is
proportional to the distance from the center. Show that its path is
a straight line through the center, and that it executes harmonic
vibration about the center whose period is independent of the starting position, and inversely as the square root of the absolute acceleration of the field.
48. If a small hole were cut through the earth along a
diameter we should have the field in the previous example. Find
the time for a particle dropped in at one end to emerge at the other,
the diameter being 8,000 miles. Also the velocity with which it
reaches the center.
49. A particle is placed on the line joining two centers, the
acceleration towards each being proportional to the distance of the
particle. Find the center of the motion, and the time of vibration,
in terms of the absolute accelerations.
50. If a particle be acted on by repulsive acceleration from
a fixed center, proportional to its distance from the center,:.nd
start with velocity in a line with the center, show that its motion
is in a straight line, and hyperbolic.
51. The earth produces an acceleration towards its center
which varies inversely as the square of the distance from the center.
A particle is projected vertically from the surface with speed T.
To what height will it rise?  Compare with the height if the acceleration were constant. Find the least value of V such that the
particle never returns to the earth.
52. A particle moves on a given curve in the earth's field.
Its arc distance from a fixed point being s, and its distance from
the center being r, at time t, show that the tangential acceleration


14




is f= —udr/ds, where u-a2g/r2, a is the radius of the earth, and
g, its acceleration at the surface. Hence from (4), if the distance
of the particle from the center is h, and its speed V, at the start,
then at time t, v2 —7V2+2a2g(r-1 —1-1).
53. Show that a particle moving in a free or constrained path
in the earth's field has a speed at any point, the same as if it had
fallen in a line with the center to that point, from a spherical surface concentric with the earth, which is the surface of zero speed
for the moving particle.
Note. Let k be the distance from the center, in falling from
which to its starting position, the particle would have acquired its
initial speed- V. By the preceding example, V2=-2a2g(h-1-k-').
Hence find v2, etc.
54. Extend the results of the two preceding examples to a
central field which produces on a particle at distance r, an acceleration F'r, (away from the center positive, and towards the center
negative).
Ans. f=F'r.dr/ds, v2    -2 (Fr —FIc), where Fr is a function
of r whose differential is Fr.dr.  Draw the spherical locus of points
at distance I7 front the center of the field, and the radius r of the
particle cuts it in the position by, falling from which to its given
position the particle would acquire its given speed v.
55. A particle P is on the inner surface of a smooth sphere
of center C, in a field of center 0, whose acceleration G on P, varies
only when the distance OP varies. Draw the small circle of the
sphere through P whose center L is on the line OC, and show that
if G and CL are alike in sign (negative towcards 0), then the
particle will describe this circle freely, and with uniform motion,
if started upon it with speed v given by
v2_(OL/OP). (LP) 2. G/CL.
RESISTANCE OF A MEDIUM
56. The free path of a particle in a. resisting medium  alone
is a straight line. In a constrained path, the path produces the
whole normal acceleration v2/R.
57. If the retardation vary as the speed, speed is destroyed
in proportion to the space described.  The motion is geometric
towards a limiting position of rest as center.
58. If the retardation vary as the square of the speed, the
reciprocal of the speed increases uniformly at a rate measured by
the absolute retardation of the medium. The speeds at successive
equidistant points on the path are in decreasing geometric progression. The distance passed over is ultimately indefinitely large,
with no limiting position of rest.
59. Find the time in which a particle with given speed will


15




be reduced to rest, the medium retarding as the square root of the
speed.
60. Consider a retardation varying as v(a+v)
61. Consider a retardation varying as v(a-+v2).
THE CONSTANT FIELD AND A MEDIUM
62. Show by dv —fdt, that the speed of a falling particle must
continually approach the speed at which the retardation of the
medium balances gravity, as a terminal value, decreasing if greater
than, increasing if less than, and uniform if equal to such terminal
value.
63. Let the medium   retard as the speed, and denote the
absolute retardation by k. Consider a particle in vertical motion
(rising or falling), starting with speed V, and passing over a
distance s in t units of time.
(a) When rising.
dv=gdt-Uds, and v-V-gt-ks
dt=-dv/(g+-kv), and ct =og.[ (g+kV) / (g+kcv)]
(b) When falling,
dv-gdt —tds, and v-V-gt-lcs,
dt-dv/(g-kv), and kti-og[(g —kV)/(g —kv)]
Note. The two cases differ only in the sign of g. Going up,
g is retardation, or -g is acceleration; going down, g is acceleration; k is retardation in either case.
(c) Show from the preceding equations that the falling particle is approaching the terminal speed g/k.
(d) In vertical motion, the speed produced by gravity in any
time is diminished by the medium in proportion to the space described during that time.
(e) A particle is projected vertically upwards with speed V;
find its greatest height H, and time T of reaching it; also the time
T' of return to the point of projection, and the speed V' of return.
Ans. kT==log(l+1kV/g), kctEI=V-gT; and by solution of
7T'. —log(l-1cV'/g), and kH — gT'-V'.
(f) A particle is projected vertically with a speed of 96 feet
per second, and returns to the point of projection in 5 seconds.
Find: the return speed, the absolute retardation of the medium,
the times of rising and falling, and the highest point reached.
Ans. 64,.246, 2.247, 2.753, 97.94.
Note. Take g-32, and there are six unknowns in the four
equations of (e) above. The two additional equations, V7=96,
T+T —5, are therefore sufficient to determine them all. V' is at


16




once found by subtracting the two values of H. Adding the values
of T,T', then gives, 5kc-log[(1+31k)/(1-2k) ]==0.
Solution is by successive trials.  Thus, the first member is
positive for c-=.2, and negative for -=.3; hence it is zero for some
value of k between.2 and.3; etc. Multiplying the equation by.4343 reduces the natural to a common logarithms, if no table of
natural logarithms is to be had.
(g) A particle is 21/4 seconds in rising, and 23/4 seconds in
returning to the point of projection. Determine the absolute retardation, etc.
(h) A particle is projected upward with its terminal speed
for the medium. Determine the height of ascent, time of ascent
and descent, and speed of return to the point of projection.
(i) One particle begins to fall from the higher extremity of
a vertical line, at the same instant in which another is projected
upward from the lower extremity with given speed. Find the time
in which they will meet.
64. Work out the cases of the preceding example when the
retardation of the medium varies as the square of the speed. Let
a-g    c/kl, the terminal speed.
65. The Motions of type (f) Further Considered.     Show
the effect of a medium retarding as the speed, on the vibrations of
a simple pendulum, and on all motions that would be harmonic in
vacuo.
Note. With the center of harmonic motion as origin, the
tangential acceleration in vacuo is expressed by -m2s. The
medium contributes 'the additional amount — lDs, by the law
stated, since Ds is the speed. Hence, in the medium
D2s+ lkDs+m2s-O.
By substituting s=-x.exp.(-1 /2kt) this reduces to
D2x+ (m2_1/4ak2) -0.
This shows thatx determines a harmonic motion, or a hyberbolic motion, or a uniform  motion, according as I2- 41/k2  is
positive, negative or zero. Also, the factor, exp( —1/kt) decreases
geometrically with the time.
66. The amplitude of a seconds pendulum is reduced from
3 to 2.6 inches in one period. Find the absolute retardation of the
medium, and the period in vacuo.
67. Two particles, connected by a fine elastic string, are
falling, with the string vertical, and extended to its natural length.
'The upper particle is suddenly stopped; find the subsequent motion
of the lower particle, the air retarding as the speed.
Ans. If n2 be the absolute tension, /k the absolute retardation,
na the speed when the upper particle is stopped,.s=aexp ( —kt) sin nt.


17




OBLIQUE AND RECTANGULAR COORDINATES
Choose any convenient axes OX, OY, OZ, in space, and let
(x,y,z) be the coordinates of the moving point in its position P,
at time t, and (x',y',z') its coordinates in the position P' at time t'.
Thus, x,y,z, will be given functions of t, and x',y',z', will be the
same functions of t'.
If we make the velocity PS the diagonal of a parallelopiped
whose edges are parallel to the axes, those edges, from P as initial
point, are called the components of PS, the one parallel to OX
being the x-component, etc. These components are expressed by
the positive or negative numbers which measure them  (positive
when in the directions of the axes). Conversely, given the components of a velocity, construct them in their lengths and directions
parallel to the axes, from the point of application P, and complete
the parallelopiped with the constructed lines as edges; then its
diagonal, PS, represents the velocity determined by them.
With rectangular axes the magnitude of the velocity is the
square root of the sum of the squares of its components.
The x-component of the resultant of two velocities is the sum
of the x-components of the velocities, and similarly for the other
components.
If two velocities are parallel, then corresponding components
are proportional, and in the ratio of the velocities (opposite
velocities in a negative ratio).
These are simply geometrical laws of displacements, and hold
for all quantities which are properly represented by displacements,
as velocities, accelerations, etc.  A line representing any such
quantities is called a. vector.
We proceed to find the components of the velocity and acceleration of a moving point, in terms of the functions, x, y, z,
which fix its position at any time t, from their definitions as limits
of average values.
Since [x,y,z] are the components of the displacement
OP, and [x',y',z], of OP', the  components   of PP' will be
[x'-x, y'-y, z'-z], and the components of average velocity PS',
if 1/(t' —t)-n, will be [n(x'-x),n(y'-y), n(z'-z)]. Hence:
(5) The components of the velocity at time t are [Dx, Dy, Dz].
These components are functions of t which we denote for the
moment by [u,v,w], and by [u',v',w'], at time t', (respectively the
components of OH, OH', fig. 1). Thus, the components of HIF
are [u'-u, v'-v, w'-w], and those of the average acceleration
lIT', are [n(u'-u), n(v-v, n(w'-vw) ], where n=l/(t'-t);
and consequently their limits are [Du,Dv,Dw] the components


18




of the acceleration HT. Hence, replacing u, v, w, by their values,
Dx, Dy, Dz:
(6)  The components of acceleration at time t are, D2x,
D2y, D2z.
AREAL SPEED AND VELOCITY
We consider the conical area generated by the "radius vector"
of the moving point, and called the radius vector area. If U denote
that area OAP, at time t (fig. 2), which becomes U'=area OAP'
at time t', then U'-U is the areal increment in time t'-t;
(U'-U)/(t'-t) is the average areal speed; and its limit DU, is
the areal speed, at time t.
On the other hand, consider the areal displacement represented by the.plane triangle OPP', and call OPP'/(t'-t) =OPS',
the average areal velocity, from time t to time t', and its limit,
OPS, the areal velocity at time t. Since the conical and plane
areas OPP', OPP', have a limiting ratio of unity, as P' approaches
coincidence with P, the areal velocity and speed have the sanme
magnitude. Also, the areal velocity OPS, is a tangent plane to the
conical surface, along the element OP.
COMPONENTS OF AREAL VELOCITY (Rectangular Axes)
The xy-component of the areal velocity OPS, is its projection
on the xy plane by lines parallel to OZ.  Since (x, y, z) and
(x+Dx,y-Dy,z+Dz) are the coordinates of P, S, the coordinates
of their projections P",S", are (x,y) and (x+Dx,y+Dy); and the
area of the projection OP"S" is consequently by plane analytics,
/2(xDy-yDx). With oblique axes, the additional factor sinXOY
is required. Similarly for the yz and zx projections. Hence,
(7) 1/2(yDz-zDy), l9/(zDx-xDz), 1/2(xDy —yDx)
are the yz, zx, xy, components of areal velocity at time t.
By the differentiation D, of these components, we obtain the
components of areal acceleration,
(8)   (yDz —zDy), etc.
MOMENTS ABOUT A POINT.
Consider any vector PS, whose components are (X, Y, Z), and
point of application, P —(x,y,z). Then, the moment of PS about
any point 0, is double the triangle OPS, considered as to plane
OPS direction PS about 0, and magnitude, the area 2OPS. Consequently, the moment is not changed by transfer of P to any point
of the line of action, PS. Its components are defined and found
as in areal velocities, and are (in rectangular axes)
yZ-zY, zX-xZ, xY-yX, on the yz, zx, xy, planes.




In particular, the moment of velocity is double the areal
velocity, and the moment of acceleration is double the areal acceleration.
The resultant of two moments 2OPS, 2OPS' is defined as the
moment 20PR, where PR is the resultant of PS and PS'
The components of the resultant of two moments are the sums
of the corresponding components of the moments.
For, y(Z+Z')-z(Y+Y')-(yZ-zY)+(yZ'-zY') etc.
MOMENTS ABOUT AN AXIS
The moment of a vector PS about an axis OZ, is defined as
the orthogonal projection of its moment 20PS on a plane perpendicular to OZ.
Thus with rectangular axes, the yz, zx, and xy components of
the moment of PS about 0, are also the moments of PS about the:axes of x, y, and z, respectively.
With rectangular axes, construct the line OL whose axial components are the moments of PS about the axes, area measure in
the latter being taken as length measure in the former, and yz components of area as xcomponent of length, etc.
In rectangular axes, the sum of the squares of the components
of an area equals the square of the area. Since the length of the
line is found in the same way from its components, therefore: The
measures of the length OL, and the area 20PS, are equal, i. e., the
magnitude of OL represents the magnitude of the moment 2OPS.
Also, with rectangular axes, two lines are perpendicular when
the sum of the products of corresponding components is zero. Hence
-show that OP, whose components are (x,y,z), and PS, whose components are (X,Y,Z), are each perpendicular to OL, whose components are (yZ-zY, etc.)    Thus, OL is perpendicular to the
plane, OPS, of the moment.    It will be named the axis of the
moment.
Finally, the angular direction of a moment, clockwise or
counter-clockwise, as seen from the end of its axis is the same for
all moments. For, conceive P and S to move continuously in any
way so that the angle POS does not vanish; then the components
of the moment 20PS, change continuously.     Since these components and the components of the axis OL, have the same
measues, therefore L  also moves continuously.   Remembering
further, that OL always remains perpendicular to the plane OPS
as the motion of P and S continues, and that the angle POS does
not vanish, so that OL does not vanish, it is plain that the aspect
'of the angle POS from L cannot change, but remains permanently
'clockwise, or counter-clockwise, in  all positions.  By taking
(1,0,0), (0,1,0) for P and S, we find L    (0,0, 1). Thus the


20




aspect of any moment from the end of its axis is the aspect of the
angle XOY from a point on OZ. The axes are said to be in righthand order, or left-hand order, in space, according as this aspect
is counter-clockwise or clockwise.
The axis OL of a moment 2OPS, determines it in all essential
particulars, magnitude, plane, and angular direction.  Further,
considering moments about one point 0, the axis of the resultant
of two moments is the vector resultant of their axes. For the area
components of a moment, and the length components of its axis,
are identical measures in rectangular coordinates, and resultants
of moments are found by adding corresponding components, as are
also the resultants of vectors.
The axes of moments are therefore vectors, and are called
moment-vectors.  Areal velocities and accelerations are similarly
represented by vectors, whose components in rectangular axes are
in length what the area components are in area. Their x-components are the yz-components of the areas, etc., they are perpendicular to the planes of the areas and represent their magnitudes,
and their angular directions as in moments.
EXAMPLES II
1. The differential and integral equations of uniform motion
in a straight line are
D2[x, y, z]=[O, 0, 0];D[x, y, z]-[a, a', a"].
(x, y, z)=(at+b, a't+b', a"t+b")
2. The equations of free motion in a constant field are,
D2[x, y,z] —[a,a', a]; D[x, y, z] —[at+b,a't+b',
a"t+b"];x-1/2at2+bt+c, etc.
3. Show that the differential equations of free motion in a
central field, with the center as origin, are:
D2 [X, [, y, Z]-=(F/) [x, y, z]
where r is the distance from the center to the point (x, y, z), and
F is the repulsive acceleration of the center (a function of x,y,
and z, which must be negative at points where the center produces
attractive acceleration).
Note. The acceleration at P is parallel to OP, since 0 is the
center; and its ratio to OP is as F to r, where F is negative if the
acceleration is towards 0.  Also, corresponding components of
parallel vectors are in the ratio of the vectors.
5. Find the. coordinate equations which determine the path
in ex. 1. Find the areal velocity, and show from it that the radius
vector of the moving point describes equal areas in equal times in a
fixed plane. What does it signify when the vector of areal velocity
is parallel to a fixed line?


21




6. With the origin at the starting position of the particle in
ex. 2, find the areal velocity, and show thence that the path is a
plane curve, and that the area described by the radius vector varies
as the cube of the time.
7. Show that the free path of a particle in a constant field
lies in the fixed plane which contains the velocity and acceleration
at the beginning of the motion.
Note. Take the fixed plane as XOY, and show successively,
that D2z, Dz, z are each zero at any time t. Or, conceive the particle
constrained by contiguous planes to lie in the plane named, and
show that these planes produce no accelerative constraint at any
time, and may thus be removed without affecting the motion. Or,
find the vector areal velocity from the general equations (ex 2) and
show that it is parallel to a fixed line.
8. If we take the axis OX in the direction of the initial
velocity (a), and the axis OY in the direction of the acceleration
(g) of a constant field, find the differential equations of free
motion, and thence find x-12gt2, y-at. By eliminating t we find
the equation of the path, y'-2a2x/g. Interpret this motion as the
resultant of uniform motion in a straight line, and uniformly increasing vertical motion. Also show that the path is a parabola,
with axis downward, and that the speed of the particle at any time
is the speed it would acquire by falling to its position from the
directrix.
Note. By analytical geometry, y —4a'x is the equation of a
parabola, whose axis is in the direction OX, with the origin on the
curve a.t distance a' from focus and directrix and with the axis OY
a tangent.
9. A particle is started with a speed of 40 feet per second,
at a slope of 3/4 to the vertical. Find the directrix and focus of
the path. Construct the flight of the particle at intervals of onefourth of a second for the first two seconds. Draw the hodograph
of the flight, showing the velocities at the same time.
10. The same as ex 9, the slope to the vertical being 4/3.
11. Plot the points of the speed curves of the preceding motions at the times named; also find their equations.
12. Find the horizontal range for a particle projected with
given speed a, at an angle of depression p from the vertical and the
depression for maximum range.
Note. The time of flight is found by the condition that in
that time the height to which the initial velocity has taken the
particle is equal to the distance fallen by gravitation.
For answer take 0 a right angle in the next problem.
22




13. In the preceding example, find the range on a plane of
depression 0 from the vertical and the maximum range.
Ans. r= —2a2sin (p —) sin p/gsin20.
For maximum     range p —/20, r-a2/g(l+-cosO), the polar
equation of the curve which bounds the greatest distance that can
be reached in any direction with initial speed a, (a parabola with
the starting point as focus, axis downward, and highest point the
height of vertical flight).  It is the envelope of the paths of all
particles starting from the origin with the same speed in the same
vertical plane.
15. A buys a flywheel of B 16 feet in diameter, with the
guarantee that it will run safely 80 revolutions per minute. The
wheel burst and a piece.was found at a distance of 1,000 feet. A
sues B for damages; should he recover?
16.  Show that for a free path in a central field, with the
center as origin, the areal acceleration is zero, the areal velocity,
constant; and that consequently the free path of a particle in a
central field lies in a fixed plane containing the center, and the
radius vector from the center to the moving point describes equal
areas in equal times. Show also the converse, that if any free path
be a plane curve through a fixed point, and the radius from that
point to the particle describes equal areas in equal times, then the
field is central, the fixed point the center.
17. For a free path in any field show that the radial acceleration at time t is, in rectangular coordinates, D'x. (x/r) +
D2y. (y/r) +D2z. (/,r).
Note.  The radial acceleration is the orthogonal projection of
the acceleration upon the radius vector OP, and is therefore the
sum of the orthogonal projections of its axial components. The
other component of acceleration, perpendicular to the radius vector, is called the transverse acceleration.  Its component on the
x-axis is found by subtracting the x-component of radial acceleration from D2x, and is 2(Y"z-Z"y)/r2, where X", Y",Z", are the
components of areal acceleration.
18. Kepler announced, as the result of astronomical observations, covering many years, that the path of a planet is an
ellipse, with the sun at a focus; that the radius vector from  the
sun to a planet describes equal areas in equal times; and that for
different planets, the cubes of the major radii of their orbits are as
the squares of their times of revolution. Newton thence proved:
(a) the acceleration on a planet forms a central field with the sun
as center; (b) towards the sun and inversely as the square of the
distance; (c) absolute acceleration the same for all planets.. Prove
the same in order, (a) from the plane path and law of areal speed,


23




(b) from the form of the path and situation of the sun, (c) from
the relation between periodic times and radii of orbits.
Note. Newton did not announce the universal law of gravitation at that time, because, owing to erroneous' data regarding the
earth, it appeared to fail in the case of the motion of the moon.
When eighteen years later more accurate data verified the law, he
immediately made the announcement.
Outline of Proof.  (a) Since the plane and magnitude of the
areal velocity are constant, its components are constant, and the
components of areal acceleration are zero. Thus, with the sun as
origin, the components of acceleration (D2x, etc.) are proportional
to (x, y, z) (Q.E.D.)
(b) Take the plane of the orbit as the plane of xy (fig. 1),
the major axis as x-axis, and the perpendicular line through the
focus F occupied by the sun, as axis of y; P-=(x, y) is the position
of the planet at time t, and FP is r. Let h be the area described
by FP in unit time. Then, e being the excentricity of the orbit, and
21 the latus rectum,
(1) x2+-y2'rr. (2) r=-zl-ex, by geometry of the ellipse, and
from areas, (3) xDy-yDx=2h.
Differentiate (1), substitute the value of Dr from (2), and
solve for Dx, Dy, with (3), making use of (1) and (2) in the
result. Thus, if c=2h/l (4) Dx= —cy/r, (5)Dy=   c(e+x/r).
Now   differentiate (5), and after reduction by preceding
equations, we find F==rD2y/y=-2ch/r2, Q.E.D.
(c) Let the major and minor axes be 2a, 2b. The positive
value of F when r=l, or the absolute acceleration of the sun's
field on the given planet P, is ktC2ch, or 4a(h/b)2, since — =b2/a;
and the time of revolution is T —7rab/h.    Eliminating h/b,
k-=4 -2a3/T2, which is the same for all planets.
19. Show that the hodograph of the motion of a planet is a
circle.
Note. Take axes through 0 (fig. 1) parallel to the preceding
axes through F. The coordinates of the point H corresponding to
the planet P are then x'==Dx, y'=Dy. Solve(4) and (5) above for
x/r, y/r, and substitute in (1). We thus obtain,
x'2-(y'-ec)2   cc, a circle, center (0, ec), radius, c.
20. If the path of a comet is a parabola or hyperbola, with
the sun as focus, and the radius vector from the sun to the comet
describes equal areas in equal times, determine the law of acceleration, and the hodograph.
Note. The work is the same as (a) (.) above, with e equal
to or greater than unity. The hodograph is a circle, with 0 on or
outside its circumference. If the hyperbolic path is the branch
about the empty focus, then, r==ex —, which is equivalent to
24




changing the signs of e and I, and therefore of c, in the above work.
The field is then repulsive and inversely as the square of the
distance, and the hodograph is a circle.
21. A planet whose radius vector from the sun is d, is projected at right angles to that radius with speed V; determine the
excentricity of its orbit, its major and minor axes, and the period
of revolution.
Note. Vd-2h, from the areal speed, V= c(1+e) from the
hodograph, and k is 2ch.
Ans. e+l-V2d/lk, a-kd/(2k ---V3d), etc.
22. If the above orbit is a circle, find the initial velocity,
center, and character of the motion. When is the orbit a parabola?
an hyperbola?
23. A particle is started horizontally with speed just sufficient to make it go round the earth in vacuo; find the time of
revolution.
24. Find the absolute acceleration such that a particle will
describe a circular orbit of unit radius in one unit of time. Find
the time of revolution if the absolute acceleration is unity.
25. Find the velocity with which a particle must be started
in any central field whose acceleration varies as a function of r,
in order that its orbit may be a. circle, and determine the character
of the motion, and the time of revolution.
26. Given that the moon revolves in a circular path about
the earth in 28 days, find its distance from the center of the earth.
Ans. Between 29 and 30 diameters of the'earth.
27. A fine elastic string, fastened at one end, and just reaching without stretching, a smooth fixed ring G, is passed through
the ring and attached to a particle. The constant field acting, the
particle is found to rest in equilibrium at a point 0 below G; it is
then drawn aside to a given point A, and projected with a given
velocity AC. To find its position P at any time t, its path, and the
character of its motion.
Note. Let c be the absolute acceleration of the string on the
particle; then at 0, its acceleration is c.OG, and c.GO is the acceleration of the field, since there is equilibrium at 0. The accelerations on the particle in any position P, are therefore, c.PG,
by the string, and c.GO, by the field, whose resultant is c.PO.
Thus, the resultant field is central, towards the center 0, and its
acceleration is cr at distance r. Consequently (ex. 16) the motion
of the particle is in a fixed plane through 0, which must be the
plane OA C containing the initial velocity. Draw OB parallel to
AC; and take OA, OB as axes ofx andy, and P       (x, y). Hence
(ex3),D2x=-cx, D2y=-cy.


25




If OA=a, AC-b /c, the solution is
x=acos (c.t), y=bsin((Vc.t),
and for the path, (x/a))2+ (y/b) 2=1, the equation of an
ellipse referred to semi-conjugate diameters OA==a, OB=b.
In another plane through OA take OB' perpendicular to OA
as a new axis of y'; and consider the point P', whose coordinates are
x=acos(Vc.t),y'==asin(\/c.t). Show that as t varies P' describes
a circle AB', uniformly c radians per unit time, as P describes its
elliptic path AB, and that P'P is always parallel to the fixed line
B'B. The motion of the particle is therefore, the parallel projection of uniform  circular motion whose period of revolution is
27r//c. This motion of P is called elliptic harmonic motion; it is
not, however, a harmonic motion on the ellipse. Harmonic motion
in a straight line is called simple harmonic motion.
28. A particle is projected from the origin with an initial
velocity whose components are a horizontal, and b vertical, and the
same component are u, v, at time t, when the particle is at (x, y).
The retardation of the air varies as the speed (its absolute value
being k), and gravity also acts. Find equations for u, v, x, y, in
terms of t. Also equations for the horizontal range and time of
flight, and the equations for maximum  range, when the initial
speed is given.
RESULTANT MOTIONS
The general definition of the resultant of two motions is that,
if P. Q. be the moving points of the component motions at any time,
and 0 is a fixed point, then completing the parallelogram OPRQ,
R is the moving point of the resultant motion as to the origin 0.
Conceive a space carried by the moving point P, but with directions
unchanged; the point R is then carried with P in this space, and
has besides, the motion round P which Q has round 0. The motion
of Q round 0 is called the relative motion of R as to P, i. e.,
OQ=PR in length and direction at every instant of time.
The velocity and acceleration of the resultant point R above
are the resultants of the velocities and accelerations of P and Q,
since, as displacements, OR=OP+OQ, and hence for velocities,
D.OR=D.OP+D.OQ, etc.
EXAMPLES III
1. Show that a translation of the origin 0 to a new position.0' translates the resultant of two given motions the opposite
amount, 0'0.
2. Show that the parallel translation of a component motion.affects the resultant motion by the same translation.
3. The resultant of two simple harmonic motions of the


26




same period and direction is a simple harmonic motion. Also find
the resultant amplitude and phase.
4. The resultant of two or more simple harmonic motions of
the same period and phase, on different straight lines, is simple
harmonic motion of the same period and phase.
5. The resultant of two or more harmonic motions of the
same period on different lines is elliptic harmonic motion (becoming in special cases uniform  circular or simple harmonic
motion).
6. Show that a simple harmonic motion is the resultant of
two uniform circular motions of opposite directions.
7. Show that the resultant of two simple harmonic motions
in rectangular directions, of equal phase, and periods as 1:2, is a
parabolic vibration.
Note.  x —a cos 2nt, y —b cos nt, and eliminating t,
(x/a) +  — 2 (y/b)2.
8. In the same case, if there is a difference of phase, the
resultant path is a figure 8 curve.
9. Construct the path of the motion
(x, y, z) -(4cosnt, 4sin nt, 3cos 2nt).
Note. Take    a  sheet of paper, 8    units  by  87  units:and with the lengthwise middle line as axis of x, and an edge perpendicular to it as axis of y, construct the harmonic curve y=
3cos/2x, from x zero to its extreme value, 87r. Fold the paper into
a cylinder of radius 4 units, and the curve upon it takes the form
of the required path.
10. If the surface of the above cylinder be turned about its
axis (the axis of z) through the angle -b, the axes of x, y, z, not
moving, the curve drawn becomes the path of the motion,
(x, y, z)-=[4cos(nt+b, 4sin(nt+b),3cos 2nt]
11. Show that the orthogonal projections of the motions in
exs. 9 and 10 on the plane ofxz are the motions given by the x and z
coordinates alone.
12. Find the resultant of two uniform circular motions of
the same period and plane. When is their resultant a simple harmonic motion?
13.  The curve of pursuit.  A point'in a plane moves uniformly in a straight line (the axis of x). A second point moves
in the plane so that its direction of motion is always towards the
first point. Assuming that its line of motion in the beginning is
the axis of y, find the differential and integral equations of its
subsequent motion.
14. A hound sights a fox 100 yards due south.     The fox
runs due east, and is overtaken in 300 yards.   Show that the
hounds runs 354.14 yards.
27




Note. If the hound run n times as fast as the fox, and r be
the distance between them  at time t, show that ds-+di=Xdy/n,
integrate, and solve for n.
PLANE POLAR COORDINATES
Consider a point moving on a plane curve APP' (fig. 3),
whose position is P at time t, etc. Let 0 be the pole, and OX, the
the initial line, and withO as center, draw a circle of unit radius,
cutting OX at L, OP at Q, etc. Then the polar coordinates of P
are 0-arc LQ, and r=OP/OQ. These are the coordinates of the
moving point at time t, and are functions of t, which become
/ ---arc LQ', r' —OP'/OQ' at time t'.
Let QR be a positive quadrant on the unit circle. Then OQ,
OR, are respectively radial and transverse directions at time t,
and all vectors at that instant, may be expressed by their components in these directions, called their radial and transverse components. Thus, the radial and transverse components of the velocity, PT, at time t, are PS and ST, where PS is a continuation
of OP, and OST is a right angle. We proceed to determine the
radial and transverse components of velocity and acceleration.
The average velocity from time t to time t' is represented
by the vector nPP' (using n for 1/(t'-t) here and in what immediately follows). In the vector addition and substraction, nPP'
is thesame as n(OP'-OP). or n(r'OQ'-rOQ). Combining with
this the   null system, -nrOQ', +nrOQ', and      noting  that
OQ'-OQ=QQ', we obtain, nPP'-n(r'-r)OQ'+r.nQQ'.
The limit of this average velocity is the velocity PT; also,
the limit of nQQ' is the velocity of Q on the unit circle, or DO.OR.
Thus, PT-Dr.OQ+rDO.OR; and since we express the units of
direction, OQ, OR, by the terms radial and transverse, we have,
(9) (Dr, rDO) are the radial and transverse components of
velocity.
Note. If u, v, stand for the units OQ, OR, the preceding
work is the differentiation of the vector displacement OP=-ru, viz.,
D(ru)==Dr.un+rDu-=Dr.u+rDO.v (since Du is seen to be
DO.v)
The acceleration is found by a similar differentiation of the
velocity, allowing for the variation of both units, u, v, in the above
note; also, observe that Dv is the velocity of R, or -DO.u  Thus
let q stand for rDO, and the acceleration is, after collecting coefficients of u. v:
D (Dr.u+q.v)   (D2r —qDO)u+- (Dr.DO+Dq)v.
If we multiply the coefficient of v by r, it becomes Dr.q+rDq,
i. e., D(rq). Thus finally,


28




(10)  F==D2r-r(DO)2, is the radial component of acceleration.
(11).G=r-1D(r2DO) is the transverse component of acceleration.
The triangle OPT is the areal velocity, and if U denote
the radius vector area OAP, v the speed Ds, and p the perpendicular ON, on PT, the magnitude of this area, or the areal speed is
expressed in the forms,
(12)  DU —/2r'2DO    2pv.
Note. v is found in polar coordinates by the right triangle
PST.
(13)  vdv=fds= —Fdr+Grd6 by projecting F and G upon PT,
to find f.
Polar coordinates are particularly useful when considering
plane motions in central fields. Since the whole acceleration is
radial, this gives at once G=O, i. e., D (r2DO) =0. Hence,
(14)  r2DO=pv —  H, a constant, (in a central field).
This expresses, of course, that the radius vector  describes,equal areas in equal times, as heretofore shown [DU==-H, U=
JHt+c]. By this equaticn, which is dt=r2d0/H, we may eliminate dt from the differential equations.. Let a small cap D stand
for d/dO; Then D become everywhere (Hl/r2)D.
i. e., Dr=(H/r2) Mr=-HD(1/r), etc. The most important.equations thus holding for a central field, with r and 0 the only
variables involved are,
(10')  F —    (Hu)2(D2U+U), (u standing for 1/r)
(12')  pv=-H, whence vdv=(H/p)d(H/p).
(13')  vdv=fds=Fdr.
EXAMPLES IV
1. Find the laws of the central fields in which a particle will
describe the following curves, and determine the starting conditions that it may do so, in particular the curve of zero speed.
(a)  The hyperbolic spiral, r==a.
(b)  The equiangular spiral p=r sin c. [(12') and (13')]
(c)  The logarithmic spiral log (r/a)=0/c.   [The same as
(b)].
(d) The spiral in which r/a is the n'th power of sec (0/n).
(e)  r=a sec nO, and r ---a sech nO.
(f)  The spiral of Archimedes, r=a0, and the rose, r=-asin20
(f)  The. cissoid r=asinOtan0; the lemniscate (r/a) 2
eos20.


29




(g) A circle about a point on its circumference; a parabola
about its vertex; about the intersection of axis and directrix.
(h)  A conic section about a focus, 1/r — +  ecosO.
(i)  An ellipse or hyperbola about its center.
(j)  An equilateral hyperbola about its center, (r/a)-2sec20..
2. A particle describes an ellipse, its acceleration being always perpendicular to one axis; find the law of acceleration.
3. What would be the path of a projectile if gravity varied
inversely as the cube of the distance from a horizontal plane?
4. If the acceleration varied as the distance in the preceding example?
5. Determine the path when the law is inversely as the square,
of the distance.
6. Consider the laws inversely as the cube of the distance,.
and inversely as the distance.
7. If the hodograph is a circle, and the field central, the law
must be inversely as the square of the distance.
8. The speed varies inversely as the distance from the center;.
determine the law of the field, and the path.
9. A planet is projected with the speed for. a circular path
at its given distance, but at an angle of 45 degrees to its radius
vector. Show that the point of projection is the end of a minor'
axis of its orbit, and its axes are as /2 to 1.
10. A particle, acted upon by two centers, moves with constant speed, and the product of its distances from the centers is
constant. If one center vary as the distance show that the other
varies as its inverse cube.
11. Show that the radius of the sphere of zero speed for a
planet is twice the major radius of its orbit, or that ~v2=-k/r —1/2a
12. Consider the motion of a particle in a field of two centers, each varying as the distance.
13. A particle moves in one plane, and its acceleration is
constant in magnitude, but revolves uniformly in direction. Find
the motion and path described, starting from rest.
Ans. With the x as the initial position and the x axis as the'
initial direction of acceleration; n, the angular speed of the latter,
and an2, its magnitude, the motion is x=aversnt, y=a (nt-sinnt),.
the motion of a point on a circumference of radius a, rolling unifornrly3 on the y axis.
14. A particle describes the cycloid x=averso, ya(-+sin4), in a fiield whose acceleration is always parallel to the base; determine its law.
Ans. The base is the line x —2a. If b is the constant ve —


30




locity parallel to the axis, the acceleration is -b2/asinpverso, in
the direction OY.
15. If a particle be acted on by a vertical acceleration so as
to describe the common catenary y-acosh(x/a), find the acceleration and speed at any point.
Ans. C2y, cy, where c is the ratio of the constant horizontal
velocity to the radius a.
16. A particle is moving in a plane, and its acceleration is
always perpendicular to the line joining it to a fixed point of the
plane. To find the law of the field in order that the angular velocity of the particle about the point may be constant, and to determine the path.
Ans.  O=-nt, and F is zero; whence
r=-aexp(nt) +bexp(- nt), G=2nDr.
17. Show that any number of given centers, whose accelerations vary as the distance, are equivalent to a single center of the
same law.
Note. Let 1, m,.... be absolute accelerations of centers A.
B,.... positive or negative as repulsive or attractive; the resultant
on P is lAP+mBP+..., and lAE+mBE+...             0, where E is
the position of equilibrium. Taking this null system from the
resultant, it becomes, (l-m+....)EP.
18. Show that if the algebraic sum of the absolute acceleration above is zero, the resultant field is constant.
19. An attractive center which produces acceleration in proportion to the distance, moves uniformly in a straight line in a
given plane. To determine the motion of a particle in the plane
whose initial circumstances are given.
20. Two particles P, P' produce attractive accelerations, each
on the other towards itself, and varying as the distance between
them. With any initial circumstances, describe their subsequent
motion.
Ans. Call the absolute acceleration of each its mass. Then
their center of gravity moves uniformly in a straight line. Each
revolves in elliptic harmonic motion relative to the other as center'
with a mass the sum of their masses.
21. The same as the preceding, except varying inversly as
the square of the distance.
Ans. Their center of gravity moves uniformly in a straight
line. Each revolves in a conic section relative to the other as focus,
with mass the sum of their masses.
22. The same as ex. 20, except varying as any given function
of the distance, with different absolute values.
23. Every point of a plane of infinite extent produces ac

31




celeration on a particle towards itself, of absolute value k per unit
area, and inversely as the square of the distance. Determine the
field produced.
24. Using ex. 21 to modify Kepler's third law, show that the
mass of the sun is approximately 356,420 times the mass of the
earth, the sun's distance from us being 400 times the moon's, the
earth's mass 75 times the moon's, and its periodic time 13.4 times
that of the moon.
25. In an orbit in any central field, draw lines paralell to
the tangents at two points of the orbit at distances inward and
inversely proportional to the speeds at the points of contact. Show
that the line joining the intersection of the tangents to the intersection of their parallels passes through the center of the field
(use pv==H).
26. If a central orbit pass through the vertices of a triangle,
and the velocities at them  be parallel and proportional to the
opposite sides, prove the center is the center of gravity of the triangle.
THE HALF SQUARE OF SPEED.
It will be convenient here to call the half square of the speed
of a particle its VIS VIVA. Vis viva is a term that has been
used in various senses by writers on dynamics, any of which make it
some multiple of the square of the speed.
If we integrate vdv —fds between given points of a given path,
we obtain 1/v2-1/2V2=-Sfds, where V is the beginning, and v, the
ending speed, and the integral is taken over the path. In a field,
f will be determined at every point of the path, whether the motion
over the path be free or constrained, whatever the initial speed
may have been, and whenever the path be described; i. e., f is a
function of s as a geometric coordinate of the path. Hence,
The increase of vis. viva produced by a field depends only upon
the path described.
A field is called a conservative field if, still further, the increase of vis viva depend only upon the beginning and ending
points, by whatever path motion takes place from one to the other.
If (X, Y,Z) be the components of the field at the point
(x, y, z),then Xdx/ds is the projection of X upon the tangent to
the path, etc., and adding the projections of each component,
fds=Xdx+Ydy+Zdz.
The condition that a field is conservative is that the above differential is a perfect differential (so that its integral may be inde

32




pendent of the path). In other words, the components of the field
must be the partial derivatives, as to x, y, and z, of a given function,
say W==F(x, y, z), so that we have fds-dW. Thus, if the particle
starts at (a, b, c), and take any path which ends in (x, y,z), its
increase of vis viva will be F(x, y, z)-F(a, b, c).
POTENTIAL.
Conversely, consider any given function, ( (x, y, z), and let it
define a field whose components at (x, y, z) are the negatives of its
corresponding partial derivatives; and let  (x, y, z) be called the
potential of the field at (x, y, z).
The field is conservative, because ( is simply -F above; also,
vdv+df (x, y, z)=O or, 1/2v2+(x, y, z)=constant.
In wods, the sum of the vis viva and potential of a particle
moving in a conservative field, is a constant.
A potential surfaceis one, every point of which is at the same
potential. If a be its potential, its equation is 4 (x, y, z) -a. Since
a particle will move on a. potential surface with constant speed, its
tangential acceleration must be zero at every point, i. e., potential
surface is everywhere perpendicular to the lines of acceleration of
the field.
It appears that a particle in motion has on a potential surface,
a corresponding speed which it must take if it arrives at that surface.  As the potential increases, the corresponding speed decreases, until a surface of zero speed is reached; and we may say
that the particle could have acquired its actual speed by having
fallen from rest upon the surface of zero speed, to its given position.
EXAMPLES V.
1. Show that all potentials may be changed by a constant
without altering the field. Show that the resultant of two conservative fields is a conservative field, and that the resultant potential at any point is the sum of the component potentials at that
point.
2. If a point move from P-(x, y, z) to P'-(x', y', z) in a
fixed direction, lim NPP'-dq, is called its differential variation in
that direction, and lim  N  [4(xy',',z') —  (x,y,z)] -d  is the
differential variation of the function  ) in that direction.  Show
that if the function be the potential of a field, -dk/dq is the component of its acceleration in the direction q of differentiation.
3. With the axis of y upwards, show that the potential of
the constant field at (x, y, z) is gy.
4. Show that the potential of the earth's field, at distance r
from the center is ga2/r. This makes zero potential where? If a
particle fall with surface of zero speed at infinity, with what speed
33




would it reach the earth's surface? Where is its surface of zero speed
if it reach the earth with greater speed?
5. Show that any central field, varying as a function of the
distance only, is a conservative field; also the resultant of any
number of such fields.
6. If the free path of a particle which is constrained to lie
on a potential surface be called a geodetic line of the surface, prove
that the radius of curvature of a geodetic line is normal to the
surface at every point.
Note. The static definition of a geodetic line is that it is the
form assumed by a strained thread constrained to lie on the (perfectly smooth) surface; and the static proof of its fundamental
property is analagous to the above dynamical one, viz, the tension
of the thread at any point produces a resultant pressure in the
direction of its radius of curvature, which, since it is balanced by
the pressure of the surface, must be normal to the surface. Geometrically, the geodesic line between two points on a surface is a
mninimum line between the two points, and lying on the surface.
T. If a particle describe a path in any field, freely or by constraint, and its velocity be exactly reversed, it will return over the
path with velocity exactly reversed at every point.
Note. Make the free path for the moment constrained, and
show that it exerts no pressure in the reverse motion. Observe that
in the free path, v2/R must be the normal acceleration of the field.
8. A free path in a field can be described freely with only
one speed at a given point; when a particle describes it as a constrained path, the vis viva at any point differs from  the corresponding free vis viva by a constant c, and 2c/R is the normal acceleration produced by the pressure of the path.
9. A particle is constrained to move on an ellipse, with one
focus an attractive center, the other a repulsive center, the two
equal in absolute acceleration. Show that the normal acceleration
of the pressure of the path varies as the curvature.
10. Hence show that if a particle be placed at equal distances
from the above centers, it will vibrate in a semi-ellipse with the
centers as focii.
11. If a line be an axis of symmetry of a field, particles
started from a point of the axis with symmetrical velocities will
describe symmetrical paths, with symmetrical motions.     Every
free path which meets the axis at right angles is symmetrical as
to that axis.
12. A point of inflection in a free path can occur only at a
neutral point (position of zero acceleration) if the field is continuous; it can have no cusp, or other point indicating sudden


34




change in the direction of motion, except by collision or sudden
impulse.
13. An orbit in a central field is concave toward an attractive
center, convex towards a repulsive one.
Definition. An apse of an orbit in a central field is a point
*of the orbit which is at a maximum or a minimum distance from
the center. The line joining an apse to the center is an apsidal
line; the distance from the center to an apse is an apsidal distance.
14. An apsidal line is perpendicular to the orbit at its apse;
if it is an axis of symmetry of the field, it is an axis of symmetry
of the orbit (eg., when the acceleration varies as a function of the.distance only). An apse cannot be a point of inflection.
15. If a central field varies as a function of the distance
*only, then the arcs of an orbit between successive apises are equal
and similar, and subtend equal angles at the center; alternate
apsidal distances are equal so that there are not more than two
apsidal distances. The orbit is closed only if the angle between
successive apsidals is commensurable with four right angles.
16. Find the equation for determining apsidal distances.
Ans. If +(r) be the potential at distance r, (HI/r)2+2((r)
-c where c is determined by initial conditions.
17. The periodic time of a planet which is double the earth's
distance from the sun is about 1033 cays.
18. Prove from the hodograph of a planet that its velocity
can be resolved into two constant velocities, c and cc, perpendicular
respectively to the radius vector and the major axis.
19. Find the locus of the empty focus and the center of the
orbit of a planet which is projected from a given point with given'
speed.
20. Find the locus of the apses of the orbit above, in any
central field.
21. Show that there are two directions of projection of' a
planet from a given point, so that it will pass through another
given point.
II. DYNAMICS
Mass is quantity of matter. For homogeneous matter, mass is
proportional to volume; a certain volume is unit mass, and the
mass of a unit volume is density so that mass     density times
volume.
A particle is a quantity of matter concentrated in a geometrical
point; it is a fiction which is convenient and almost necessary in
the development of dynamics. Its actual function in the dynamics
of finite bodies will be considered later.


35




.. The Momentum of a particle is the product of its mass into
its velocity. It is consequently a vector quantity, represented by
the extension of the velocity vector as many times as there are
units in the mass of the particle.  Remembering that the momentum of a particle is tangential to the path, it is fully determined by the numerical measure, my, where m is the mass, and
v the speed, of the particle; but momentums add by the parallelogram law of vectors, and not by the addition of their numerical
values. Thus, the sum of two momentums of 3 and 4 units, varies
from 7 to 5 to 1, according as the angle between their directions
varies from 0 to 90 to 180 degrees.
The fundamental basis of mathematical dynamics is the postulate that a particle cannot change its own state of momentum.
Such change must come from outside causes, so that if a particle
of given mass were wholly unacted upon from without, it would
move forever uniformly in a straight line (including permanent
rest as the limiting case of zero speed).
Any cause of change of momentum in a particle is named
impulse or force, according to the character of the change.
An Impulse produces instantaneous change of momentum.
(The mathematical analogue of a blow). It has magnitude and
direction, which are represented by the change of momentum, i. e.,
the magnitude of the suddenly added momentum, and its direction,
are the determining elements of the impulse, so far as it can be
mathematically expressed.
A  Force produces gradual and continuous change of momentum, i. e., an indefinitely small change in magnitude and direction, in an indefinitely small time.
The Average Force during any time, is represented by the
quotient of the whole change of momentum by the time of change,
i. e., the average change per unit time. It has direction and magnitude as represented by this quotient.
The Force acting at an instant in the middle of a continuous
change of momentum, is the limit of the average force (as estimated for an indefinitely small time whose limit is the instant
considered). It has magnitude and direction represented by such
limit of the average force, and is expressed by components.
Impulses and Forces are plainly vector quantities, being either
differences of vector momentums, or the limits of such differences.
The entire system of dynamical quantities, velocity, momentum,
acceleration, impulse and force, rest fundamentally upon displacements in space, in which a displacement from a point A to a point
B plus another from B to C, equals a displacement from A to C.
We can take the lines which represent the velocities of the
particle in fig. 1, to represent the momentums instead, since there


36




is no change of direction, but simply a change of measures of
length. Thus, Off will be a line of mv units, and OH', a line of
m' v' units, so that my replaces v andm'v' replaces v' in the work
of finding the components of acceleration. Hence:
The Tangential Force at time t is d(mv)/dt-=D(mv).
The Normal Force at time t is mvdc/dt —mv2/R.
If m is a constant, then D(mv)-=mDv=mf. Also, the normal force is in any case, the mass times the normal acceleration.
Hence,
When the mass concerned does not vary with the time, the
force acting equals the product of the mass into the acceleration.
A further postulate of mathematical dynamics is that action
and reaction between particles are equal and opposite.
The preceding postulates and definitions were in point of fact
successively deduced from observation and experiment as true in
nature, beginning with Galileo, and ending with Newton, who first
formulated them as a basis for a mathematical science of dynamics
under what he called the three laws of motion. The most crucial
tests that these laws are in accord with nature are the many verifications of the Newtonian Dynamics in Astronomy (see Encyclopedia Britannica, article "Mechanics," sections 3 and 4).  The
theory has been applied successfully in the development of all departments of Physics and even of Chemistry, and the theory of all
natural phenomena is becoming more and more a purely dynamical
science.
The true measure of mass is dynamical. That is, mass is a
property of inertia in matter, in virtue of which external action is
required to change its state of rest or of uniform motion in a
straight line. Thus, equal masses are those which under the same
external actions, observe the same changes in velocity. This is
reduced, in practice, to the case of external actions in equilibrium.
Thus, equal masses will in vacuo extend a spring balance equal
amounts; or, placed in the opposite pans of an arm balance, will
not disturb its equilibrium; or, coming together with equal and
opposite velocities, and adhering, will be brought to rest; etc.
FINITE BODIES.
A finite body can be divided by imaginary surfaces into indefinitely small elements. The surfaces are determined      by the
coordinate system, and are such that along any surface, one of the
coordinates of the system is a constant. Thus, in the Cartesian systems, x,y, z, the surfaces are planes parallel to the coordinate
planes. In the polar system, r, 0, 4, the surfaces are spheres about
the pole as center, circular cones about the initial line as axis, and


37




planes through the initial line. In the cylindrical system, r, 0. z,
the surfaces are cylinders about the axis of z, planes through and
planes perpendicular to, that axis.
Consider the element which contains a given point P, and
ultimately reduces to that point.  So long as this element of the
body has not vanished, it has form, volume, and mass; and we may
construct a similar element in form and distribution of mass, in
so large a proportion to the vanishing element that it remains.
finite, and approaches a limit, called the differential element at P.
The limits of the sums of all the elements of volume and mass of
the body are plainly the whole volume and mass. But these are
the integrals of the corresponding differentials, by definition of
integration.
The differential element pertaining to a point P, is rigorously
a particle of mass din, concentrated at P.  In fact, its distance
from any fixed point, line, or plane, corresponds, in the summation
of the indefinitely small elements of the body, to the distance of
some point of the small element to which it corresponds, and all of
these points become in the limit, the single point P. Also, in considering the forces on a body, they reduce to the systems in which
the forces on each element are those which produce its actual
motion as a separate body, and in proportionalizing the forces on
an element with the element, they become ultimately forces acting
at P, whose resultant is the differential force on the particle dn.
Thus, the rotational effects of the forces on the small elements
disappear in the limit.
The integral of particles over a line is a filament; the integral over a surface, is a lamina. Forces on the several particles
must, of course, be integrated in respect to their components in fixed
directions, and in respect to their moments about fixed axes in
order to obtain their resultant effects upon the body as a whole.
The force on a particle is its mass times its acceleration. The
mutual actions of a system of particles are called internal forces;but when the integration of a component or moment of force on a
particle extends to every particle of the system, the effects of the
internal forces mutually cancel, since action and reaction are equal
and opposite, so that the integration obtains simply the corresponding component of resultant external force, or resultant mo —
ment of external force, as the case may be.
KINETIC ENERGY AND WORK
If m be the mass, and v the speed of a particle, its kinetic energy
is 1/mv2, or the product of mass into vis viva. The kinetic energy
of a body is /2v2'dm, extending to every point of the body.


38




The work done by a constant field of force in any displacement
of the body is, force times displacement times cosine of angle between them, it may also be expressed as, force times, component of
displacement in its direction, or displacement times component of
force in its direction.
The sign of the work is positive or negative according as the
angle between force and displacement is acute or obtuse; it is zero
if the angle is a right angle.
The work done in two successive, displacements is the work
done in the resultant displacement; and the work done by the
resulant of two constant fields is the sum of the works done by
each.
The work done by a field which varies from point to point
must be considered with reference to the path. We divide the path
into elements so small that the field is nearly constant for the displacement of any element; and so rigorously constant in the differential field corresponding to the point to which the element
ultimately reduces. In other words, the exact work done is the
integral over the path of the differential work at each point-remembering that this stands for nothing else but the limit of the
sum of the works done along each element under the more and
more nearly rigorous hypothesis of the constancy of the field over
the element.
If W denote the work of the field on a particle. which describes the path AP (fig. 1) in the field, we have,dW —mfds-mvdv.
Thus, if V be the speed at A, this gives, by integration,
W —=l/ mv2- 1 mV2.
In words: The work done by a field upon a particle. movingover any path equals its increase of kcinetic energy.
This will include also the work done by a mediuii, friction,,
and any other forces, whose tangential components are included in
mf above.
This gives a new definition of the potential at P in a conservative field, viz., the work the field will do in bringing a particle
of unit mass from P to the surface of zero potential; or, the work
that must be done against the field in bringing it from zero to its
given potential.
TRANSLATION OF A RIGID BODY
In what is called translation of a body, all points describe
equal distances s, in parallel straight lines. The speed v-Ds of"
any point, is the speed of the body, andDv is its acceleration.
The momentum is 5vdm==mv. 
The kinetic energy is Sl/2dm.v2'1/2mv2.
The resultant force is Sdm.Dv=mDv. Since the forces on alli


39




the particles are parallel and proportional to their masses, this
resultant force passes through the center of gravity of the body.
ROTATION OF A RIGID BODY
In this motion, all points describe circular arcs of equal
radian measure 0 about a common axis, so that if r be the perpen-,dicular distance from  the axis to a given particle P, the arc
distance described by the particle is s-r0; r varies with the position of the particle, but not with the time; 0 varies with the time,
and is the same for all particles.  DO,= w, say, is the angular
velocity of the body, and Dw is its angular acceleration.
The velocity of the particle P at distance r from the axis is
Ds=rD0-rw, perpendicular to r.
The momentum of the particle is therefore wz.rdm, perpendicular to r, so that its moment of momentum about the axis is
w.r2dm. Let I stand for Sr2dm, extending to every particle of the
body. Then,
Moment of Momentum about the axis is Iw.
Similarly, from  /2v2dm, and v=-rw, the Kinetic Energy is
1/21W2.
The tangential force on P    is, dm.Dv=-rdm.Dw, and the
normal force is, dm.v2/r-r=dm.w2.    Since the latter intersects
the axis, its moment about the axis is zero. Thus, the moment of
the force on the particle P, about the axis is, r2dm.Dw. Hence,
The resultant Moment of Force about the axis is, IDw.
One should observe the parallel between translation and rotation. In the former, v and Dv, are linear velocity and acceleration, and mv, 1/2mv2, mDv, are the momentum, kinetic energy, and
resultant force. In the latter, w and Dw, are the angular velocity
and acceleration, and Iw, /2Iw2, IDw, are the moment of momentum, kinetic energy, and moment of force. The inertia factor m in
translation, is replaced by the inertia factor I in rotation, which is
consequently called the moment of inertia of the body about the
axis. From its definition, the moment of inertia depends solely
upon the distribution of mass about the axis, and, is made up of
the sum of the moments of inertia of its component parts. The
distance from the axis at which a body can be concentrated with
the same moment of inertia is called its radius of gyration about the,axis. Thus I7=mk2, where k is the radius of gyration.
It is shown in the calculus that if the axis, originally at the
center of gravity, be moved parallel to itself a distance a, the:square of the new radius of gyration is k2+a2.


40




THE NORMAL FORCE OF REVOLUTION.
Since the normal forces meet the axis, they are balanced by,
and determine the stress on, the axis. It is necessary to consider
coordinate axes, and moments about them, for which refer anew to
the topics on moments (page 17). Take the axis of revolution as
axis of z; let the plane of xy contain the center of gravity of the
body; and let P be any point of the body whose coordinates are
(x, y, z), mass dm, and distance PZ-r from the axis of z. (With
some constructions by the student, named portions of fig. 2 will
answer as a diagram).
Let PN (on PZ) be the normal force on P. We have shown
above that PN-=rw2dm, or in x, y, z, components,
-w2xdm, — w2ydm, 0.
Introduce the balancing forces OM, -OM, at 0, where OM=PN.
Thus, the force PN on the body is replaced by the equal and
parallel force OM, acting at 0, and the couple, PN, -OM, which
is fuly expressed by its moment 20PN. Finding the components
of this moment from the components of PN and the coordinates
of P, (Yz-Zy, etc.) they are, w2dm.yz, -w2dm.zx, 0.
Do the same for every particle of the body, and integrate the
several components of force and moment, extending to every
particle of the body. The coordinates of the center of gravity are
say, (x', y', 0), so that (m=-dm being the whole mass),
xdm=mx', Sydm=my'
Also let 5yzdm=nA, Szxdm-B. These integrals are called products
of inertia of the body about the axis; and when they are both zero,
the axis is called a principle axis of the body for the point 0.
Using these integrals, we thus resolve the normal forces of revolution into a single force acting at 0, and a couple, whose components are:
The force, -w2mx' -w2my', 0.
The couple, w2A, -w2B, 0.
Plainly, the resultant normal force is the same as if the whole
mass of the body were concentrated at its center of gravity, so that
it is zero only when the axis of revolution passes through that
center of gravity. Also, the couple is zero only when the axis is a
principle axis for the point 0. In words, there is no stress on an
axis of revolution of a body (direct or twisting) when and only
when the axis is a principal axis of the body through its center of
gravity.
THE COMPOUND PENDULUM.
Any body revolving about a fixed horizontal axis is a compound pendulum. Let OG=a, be the distance of the center of
gravity from the axis, 0 the angle between OG and the downward


41




-vertical, OL, and I, the moment of inertia about the axis. Computing moment of the force (gravity) about the axis,
I.D26 ---mga sinO.
Compare this with the equation for the simple pendulum of length
1, at the same angle, at any time t, 1.D2= —gsin0, and we find that
l=I/ma, which is called the length of the equivalent simple pendulum.
Produce OG to C, making OC==I/ma; C is called the center
of oscillation as to the center of suspension O. Let k be the radius,of gyration about a parallel axis through the center of gravity G;
then I=m (a2+ c2). Thus OC=a+-b, where
b —GC=k2/a.
Conversely, if C be made center of suspension with a parallel
axis, its center of oscillation will be on CG produced, at a point
which is a distance k2/b, or a beyond G, which makes it the point
0. Thus, centers of suspension and oscillation are interchangeable.
EXAMPLES VI
1. Prove that the whole work done in raising a system of
bodies through different heights is the same as raising the entire
weight through a height equal to that which their center of gravity
is raised.
2. Find the work performed in moving a ton 100 yards on a
uniformly rough horizontal road, the coefficient of friction being
*one-tenth.
3. Show that the same work is expended in drawing a body
up an inclined plane, subject to friction, as would be expended by
drawing it along the base and raising it vertically upward.
4. If two particles attract each other directly as the product
of their masses, and inversely as the square of the distance between them, find the work done, whet they have moved from an
infinite distance apart to the distance r.
5. Find the potential outside, upon, and within, a spherical
shell, of uniform mass per unit area. Derive thence its attraction
upon a particle in the same situations.
6. Show hence, that a solid spherical mass, whose density at
any point varies only with the distance from the center, attracts
an outside mass as if its whole mass were concentrated at the
center.
7. Determine the law of force wthin a tube along a diameter
of the earth.
8. A flat pivot presses against a rough plane; find the work
done against friction in one revolution of the pivot.
9. A ball moving with a velocity of 1,000 feet per second has


42




its velocity reduced by 100 in passing through 1.9 inches of wood;
what is its maximum penetration in the same material?
10. A rectangular log lies on the floor of a car, in direction
perpendicular to the sides. If it be prevented from slipping, determine the speed of the car, such that if suddenly stopped, the
log shall be just overturned.
11. A sling is formed by fixing the ends of an elastic cord
(natural length 2a) to the prongs of a stick. A bullet of mass m is
placed at the middle of the cord and drawn back until the cord is
of length 2b. Determine the velocity which the sling will give to
the bullet.
12. An elastic cord, absolute tension k, is attached to a mass
m, resting on the ground.   The free end of the cord is raised
vertically until the mass is just lifted; find the work done.
13. An indefinitely long chain, mass m per unit length, is
coiled on the ground. A ball of mass kcm attached to the end of
the chain is projected vertically with speed V; determine its speed
at any time and the height to which it rises. (Note. The mass is
here variable). Consider the same projected on a smooth horizontal plane.
14. A rod of uniform density is let fall from a horizontal
position, one end being attached to a fixed horizontal axis. Find
its angular velocity when vertical.
15. A weight is attached to a cord which is wound round a
homogeneous cylinder with axis horizontal. The weight falls,
turning the cylinder. Find its speed at any time, if the cord is
weightless, and also allowing for its weight.
16. Find the speed acquired by the center of a hoop in rolling
down an inclined plane of height h.
17. A homogeneous cylinder or sphere, rolls down an inclined plane without slipping. Investigate its motion.
18. Find the center of oscillation of a uniform circular plate,
with respect to any point as center of suspension. Also, the position of the center of suspension for minimum time of sriall oscillation.
19. Find the center of oscillation of a homogeneous sphere,
about a tangent, and compare its time of oscillation with that of
a spherical shell of equal diameter, about a tangent.
20. A pendulum is an elliptic disc, axes four and two feet,
point of suspension a focus; determine its period and center of
oscillation.
21. A bent lever ACB rests in equilibrium when AC is inclined at an angle e to the horizontal; show that if this arm be


43




released from a horizontal position (C being fixed.) it will fall
through the angle 2e. Find its center of oscillation.
22. A right cone oscillates round a horizontal axis through
its vertex and perpendicular to its axis. Find the length of the'
equivalent simple pendulum. If the initial position of the axis is
horizontal, find the angular velocity when it reaches a vertical
position.
23. A bullet weighing 50 grammes is fired into the center of
a target with a velocity of 500 meters a second. The target weighs
a kilogramme and is free to move. Find in Kilogramme meters
the loss of energy by impact.
24. A body of mass m sliding upon a perfectly smooth horizontal table is connected by a string passing through a smooth hole
in the table with another body m' which hangs freely; find its time
of revolution in a circular path of radius a, in order that m' may
remain at rest.
25. A uniform rod AB is held by smooth rings at AB to'
smooth horizontal and vertical intersecting rods. Investigate its
motion.
26. In a swing at rest, the horizontal bar (2a) is given an
angular velocity c about a vertical axis through its middle point;
to what height will it rise?
27. Find the kinetic energy in the above bifilar suspension
when the angular displacement is 0. Also, the time of small os;illations.
Note. If I be thelength of each cord of suspension, and. its
inclination to the vertical, then lsin —=2asin/20, and 1(1-cosO) is
the height of the rod above equilibrium.
28. If n be the torsional rigidity of a cylindrical rod, radius
b and length 1, fixed at one end, determine the couple necessary to
twist the free end through a small angle 4.
- Note. At a point (r, 0) in the free end with center as origin,
the strain is s-=r4/l and ns is the stress per unit area; thus
ns. r dr dO is the stress at the point; and since its direction is perpendicular to r, we multiply it by r to obtain the couple, whose
integral over the end is the required couple.
30. Two circular disks have a wire as a common axis, to which
they are ridigly attached. The ends of the wire are stretched in
a vertical line and fastened to rigid supports.  If each disk be
turned through a horizontal angle and released, determine the differential equation for the angular displacement of each at time t.
Determine the motions of common period which they may have
under certain starting conditions, by substituting for the angles,
values of the form acosnt, bcosnt.


44




31. Prove that neither the moment of momentum of any sys-tem of bodies about a fixed axis, nor the motion of their center of
gravity, can be changed by their mutual actions.
32. A particle moves on the surface of revolution with its.axis vertical.  Prove that its moment of momentum  round the
axis is constant.
33. The Ballistic Pendulum.    A  compound pendulum     is
struck when at rest by a bullet moving in a horizontal direction at
right angles to the axis. If I be the moment of inertia about the
axis of the imbedded bullet, m' its mass, a' the distance of its
center of gravity from the axis, and c the angle described by the
pendulum under the influence of the blow, show that the speed v,,of the bellet, is given by (m'a'v)2_
2g(1-cos) (1+1') (ma+m'a').
34. A heavy string of variable density hangs in the field of
gravity with both ends fixed. Investigate the differential equations
of the curve and tensions. (Calc. pp. 86, 87.)
35. Find the laws of density, and the tensions, when the
-forms are, (a) a parabola with axis vertical (b) a common cutenary, (c) a semi-circle, with diameter horizontal.
36. Two equal particles are connected by an inelastic cord;,one lies on a smooth table, the other just over the edge, the cord
at full stretch and at right angles to the edge. Find the time to
the instant when the whole cord becomes vertical.
-:7. A particle descends from rest down the convex side
*of the logarithmic curve, y-logx/loga, whose asymptote is horizontal. To find where it leaves the curve.
38. A particle is attached to a thread which winds once
round a circle whose center is repulsive, and varying as the distance.
'To find the time of unwinding, and the tension of the thread at
any time.
39. A rectilinear tube revolves uniformly about one extremity in a horizontal plane; to find the motion of a particle within
the tube.
38. Supposing the force which acts on the crank of a steam
*engine to be vertical, and as the sine of the angle through which
the crank has revolved from the vertical position; to find the
angular velocity of the crank in any position, the moment of re*sistance being half the greatest moment of the force, and the
moment of the weight of the crank being regarded as inconsiderable.
40. A string lying in the form of a circle on a smooth table
is revolving like a wheel; to find its tension.
41. A rod revolves freely about a fixed point at its upper


45




end so as to be inclined at a fixed angle to the vertical; to find its
angular velocity, and the direction of the pressure at the fixed
point.
42. A hemisphere oscillates about a horizontal axis which
coincides with a diameter of the base. Its base is inclined at an
angle of 60 degrees to the horizon at the beginning of the motion..
To find the pressure on the axis at any time, and the maximum
pressure. Also the center of oscillation.
43. A homogeneous sphere is suspended by a fine wire attached to a fixed point on its surface.  It is turned through n
revolutions and released.  Determine the untwieting motion, -lhe.
elasticity of torsion being proportional to the angle.
44. If a partcle be attached by a string of glvnri length to a
po(ilt in a rough horizontal plane, and a given velocity be collmlunicated to it, perpendicular to the string supposed tight, Jind the
lension of the string at any time, the time in whlich it will be
brought to rest, the whole arc described.
45. Two equal rings, P, Q, slide on a;tri nr which passes
round fixed pegs A, B, in a smooth horizontal plane. The rings
are brought together, and projected with equal velocities so as to
kl(:ep  he string stretched symmletrically.  Show that the path of'
(ec.h is a paraboil, nIld that the ten-sion of the stri.ng vaTtries inveriv as the liiauee of eaci from the nearer I; —
46. A particle moves in an elliptic tube, under the action of
a rod with its ends in the focil, each element of which attracts inversely as the square of the distance. Show that its speed is constant, and find the pressure of the tube at the end of the minor
axis.
47.  Conservation of Areas. If a system of particles be subject only to mutual actions, and the products be made of the mass
of every particle into the projection on a given plane of the area
described by its radius vector, show that the sum of those products.
varies uniformly with the time.
48. A uniform straight plank rests with its middle point on
a rough horizontal cylinder, their directions being perpendicular
to each other. The plank is slightly displaced so as to remain in
contact with the cylinder without slipping; to determine its period
of vibration.
49. An inelastic beam, one end fixed to a horizontal axis,.
falls from  a given position and impinges against an immovable
obstacle; to determine the shocks at the obstacle, and on the axis.
Also, determine the position of the obstacle in order that there may
be no shock on the axis.
NOTE:-Let R, R' be the impulsive reactions at the two places


46




which instantly destroy the motion, and take moments about the
axis, and a parallel axis through the center of gravity. The required
position, called the center of percussion, coincides with the center'
of oscillation.
50. A uniform bar, one end attached to. a horizontal axis so'
as to turn in a vertical plane, falls from a horizontal to a vertical
position, where it strikes a perfectly elastic ball. To find the position of the ball that it may receive the greatest velocity, and the
velocity.
51.  One end of a straight brittle rod is held still in the hand,.
while the other is tapped against a table till the rod snaps; to determine the point of fracture.
52. An inelastic beam, which is moving without rotation.
along a smooth horizonal plane, impinges upon a fixed vertical rod;'
to determine the impulsive reaction of the rod, and the subsequent
motion of the beam.
53. A   cylinder or sphere,radius a, mass m, moment of'
inertia. I about axis or diameter) is started with linear velocityV,.
and angular velocity W greater than V/a, about a horizontal axis
down a 'plane of inclination b, and coefficient of friction c; de —
termine its subsequent motion.
Ans.   At first, Dv-g (sinb-c.cosb), IDw ---cagcosb.     As'.
soon as aw —v, the motion changes its character.  If friction is
sufficient to cause pure rolling, and F denote the tangential reaction of the plane, we have,
v=caw, mDvrnlgsinb-F, I.Dw.
— aF; whence, F =mIgsinb/ (I-+ma2)
The largest F which the plane can produce is c.mgcosb, and if
this is less than above value, then instead of the preceding pure
rolling we have Dv=g (sinb-ccosb), I.Dw-nmagcosb; aw becomes
less than v, and when it becomes zero, the further motion is pure
sliding.,54. A sphere is laid upon a rough inclined plane; show that
it will slide or roll according as the cefficient of friction is less or'
greater than 2/7 the tangent of the inclination of the plane. IDetermine the like conditionr for a cylinder.
a5.  Two spherical shells, of the same outer dimensions,
weight, and appearance, are in. reality of different inner cavity.
Determine dynamically the thicknesses and densities of the shells..


47




S'snS
V S
N~~~~~   C\.~
FIG.2..^^/7S  FIG~3.
/          -
X
~,  ~~~%'I 0 
A.